Chris Moody wrote 7 lines
Hi Marc,
Thanks for the feedback. One more question and I think I'm done for now
and able to finish the program I'm writing:
If FindNode does not locate a node with the given name, what does it
return?
I don't know. It looks like it would be the first child of the subnode you're working at. I'm not sure if that first child can be a nil pointer, so you need to check yourself or someone else has to answer this question.
But it is easy to jump to the implementation of FindNode() yourself:
1. Put the cursor on the function call to <yourxmldoc>.FindNode()
2. klick the right mouse key
3. in the popped up menu select "Find Declaration of FindNode"
4. hit the keys Shift+down (arrow)
After Step 3 you're shown the class declaration, then after the fourth step the cursor is sitting in the implementation of the method in question. Then you can read the code.
Have fun,
Marc
Show quoted text 38 lines
Thanks,
Chris
On 07/24/2015 01:34 PM, Marc Santhoff wrote:
> On Fr, 2015-07-24 at 11:57 -0700, Chris Moody wrote:
>
>> Would it be easiest to introduce a fourth element to the XML that says
>> how many children the 3rd element has? Or is there an easy way already
>> to get this and just run a for next loop to read each child?
> Maybe sth. like this would do:
>
> <counted_element count=3>
> <child>1>/child>
> <child>2>/child>
> <child>3>/child>
> </counted_element>
>
> or similar. Besides making the count an attribute of the root node of
> counted elements you could simply use a standard pascal loop:
>
> n: XMLNode;
>
> { this is only pseude code ... }
> while (xmldoc.hasMoreChilds()) do begin
> n := getChild();
> ... process n ...
> end:
>
> For real code
http://wiki.freepascal.org/XML_Tutorial mwill help.
>
> HTH,
> Marc
>
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http://lists.freepascal.org/cgi-bin/mailman/listinfo/fpc-pascalMarc Santhoff